New answers tagged

0

From the information given we can calculate: $AB = (AB)^* = B^* A^* = B^* A$. To be able to conclude that $B^*=B$, we would need to be able to invert $A$, which would give us: $B^* = A B A^{-1}$. And then we would need that $B = A B A^{-1}$ for which I have forgotten the name.


0

Using your $g(t) = t - \frac{\pi|{2}, for\, t > 0$, and $g(t) = t +\frac{\pi}{2}\,for\, t< 0$, We have $$g(t) - t = -\frac{\pi}{2}\, for\, t > 0 $$ and $$g(t) - t = \frac{\pi}{2} \, for\, t < 0$$ So (g(t) - t) is odd function. so $$\int_{_pi}^{\pi} s(t)(g(t) - t)dt = 0 $$, for any $s \in Y$ For any $f \in Y$, we have $\int_{-\pi}^{\pi}f (g -h) ...


3

Choose $N\in \mathbb{N}$ so that $\sum_{n\geqslant N} \|f_n-e_n\|^2 < 1$. We will show that $(f_k)_{k\geqslant N}$ is a basis for $\{e_k\colon k<N\}^\perp$, which will be enough. Let $f\in \{e_k\colon k<N\}^\perp$ and suppose that $(f,f_{n})=0$ for all $n\geqslant N$. I claim that $f=0$. Assume not. Then $$ \|f\|^{2}=\sum_{n\geqslant ...


1

$i(A+1)$ is a symmetric tri-diagonal matrix on each of the subspaces spanned by the even and odd coordinates. These are equivalent to Jacobi matrices (real symmetric with positive off-diagonal elements). It is a classical result that these define essentially self-adjoint operators if the off-diagonal elements grow no more rapidly than $n$. By Theorem 2.7 of ...


1

It means that for any scalar $s\in C$ and for all vectors $f,g_1,g_2$ we have $<f,s g_1+ g_2>=\bar s\cdot <f,g_1>+<f,g_2>.$


1

If you provide more context regarding where you have met a weight function in Quantum Mechanics, someone might comment about the physical meaning. For motivation, recall that even on a finite dimensional vector space, there are many different possible inner products on a single vector space - not only the standard one. A choice of an inner product on a ...


2

If $\phi_n= \psi_1 + \psi_{n+1}$, then with $$x_n:=\sum_{k=1}^n \frac{1}{n} \phi_k =\sum_{k=1}^n\frac{\psi_1}{n}+\sum_{k=1}^n\frac{\psi_{k+1}}n= \psi_1 +\frac{1}{n}\sum^n_{k=1}\psi_{k+1}$$ you get $$\langle x_n - \psi_1, x_n - \psi_1\rangle = 1+||x_n||^2 -2 \langle x_n ,\psi_1\rangle = ||x_n||^2-1$$ Now ...


0

Let $S=\{x\in\mathcal{H}:\|x\|=1\}$. Since $S$ is closed and bounded, then $S$ is weakly compact. But $T$ is compact. Then $T$ is weak-norm continous. Thus $f:H\to\mathbb{R}$ given by $f(x)=\|Tx\|$ is weak-continous. Since $H$ equipped with the weak topology is Hausdorff and $S$ is weakly compact, then $f$ attains a globlal maxima on $S$. That is, there is ...


0

Regarding your comment about bounded functionals. Your answer is not correct. Hahn-Banach theorem works in other direction. Given $x\in X$, then there exists a bounded functional $f$ with $\|f\|=1$ and $f(x)=\|x\|.$ Your argument proves what I claimed. What you need to do on Hilbert spaces, just apply Riesz representation theorem to obtain a vector $y\in ...


3

We will prove that $\exists x_0\in S(0,1)=\{x\in H:\|x\|=1\}:\|Tx\|=\|T\|$. You know that $\|Tx\|\leq \|T\|\,\forall x\in B(0,1)\Rightarrow \exists$ a sequence $\{x_n\}\subset B(0,1): \|Tx_n\|\to \sup\limits_{x\in B(0,1)} \|Tx\|=:\|T\|$. From this sequence you can chose a weakly convergent subsequence, still denoted by $\{x_n\}$, in $B(0,1)$ , say ...


2

If you prove that the image of the closed unit ball of $\mathcal H$ by $T$ is compact in $\mathcal H$, then there exists $x\in \mathcal H$ with $\|x\|=1$ such that $\|Tx\|=\|T\|.$ Indeed, since $T(B_\mathcal H)$ is compact, the norm $\|\cdot\|$ atains its maximum on $T(B_\mathcal H)$. Therefore, the set $\{\|Tx\|:\; x\in B_\mathcal H\}$ is closed. This ...


0

In your proof slight modification is needed. We have, $UU^* =U^*U =I$. So $U$ is invertible with $U^{-1}= U^*$. Also as you have already shown that $||U || = 1 = ||U^{-1}||$, we must have $$\sigma (U) \subset \{\lambda \in \mathbb{C} : |\lambda| \leq 1\}\;\; \text{and}\;\; \sigma (U^{-1}) \subset \{\lambda \in \mathbb{C} : |\lambda| \leq 1\}.$$ Now to ...


0

You don't seem to have set up the approach properly. You should start with the tensor product of the $H_i$'s, i.e. $H=H_1\otimes H_2\otimes \cdots \otimes H_n$. This is a quotient vector space. An expression of the form $\phi_1\otimes \phi_2\otimes \cdots \otimes \phi_n$ represents an equivalence class in $H$. In order for the inner product to be ...


0

We have (by uniqueness) that $u = M_{{\mathrm sgn}\,\varphi}$, where $$ {\mathrm sgn}\,\varphi(x) = \begin{cases}0, &\varphi(x) =0,\\ \frac{\varphi(x)}{|\varphi(x)|}, &\varphi(x)\neq 0.\end{cases} $$


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No. Any operator on a finite dimensional space is compact and in particular, any non-trivial nilpotent operator $T$ is a counterexample.


4

Consider the space $H = L^2([0,1],\mu)$, where $\mu$ is the Lebesgue measure. Define $T$ to be the multiplication with the identity function, i.e. $$(Tf)(x) = x\cdot f(x).$$ Since the identity function is bounded, $T$ is bounded ($\lVert T\rVert \leqslant 1$), and since it is real-valued, $T$ is self-adjoint. Clearly $T$ has no eigenvalues, since $$(T - ...


1

It occurred to me that your problem has to do with a creation and annihilation operator, according to \begin{eqnarray*} X &=&U+V \\ a^{\ast } &=&V,\;a=U \end{eqnarray*} see below. Let $\mathcal{H}=l^{2}$ \ with elements $u=u_{1},u_{2},\cdots $ and let $K$ be defined by \begin{eqnarray*} \mathcal{D}(K) ...


1

Your conjecture is false. It is always the case that $T^{\star}T$ is densely-defined and selfadjoint if $T$ is a closed densely-defined linear operator on a Hilbert Space $H$. Let $H=L^2[0,1]$ and let $\mathcal{AC}[0,1]$ be the absolutely continuous functions on $[0,1]$. Define $T=\frac{d}{dt}$ on the domain $$ \mathcal{D}(T)=\{ f \in \mathcal{AC}[0,1] ...


3

Well, no answer has been given, so here's one: We know $\{\cos nx:n=0,1,\dots \}\cup \{\sin nx:n=1,2,\dots \}$ is an orthogonal basis of $L^2[-\pi,\pi].$ Suppose $g\in L^2[-\pi,\pi]$ is odd. Then $$\int_{-\pi}^\pi g(x)\cos nx\, dx = 0, n=0,1,\dots $$ Thus such a $g$ can be written uniquely as $$ \tag 1 g(x)=\sum_{n=1}^{\infty} b_n \sin nx,$$ the sum ...


2

Even with the correction I gave in the comment, this just isn't true: Consider $H = \Bbb R^2, V = \{(r, 0) \mid x \in \Bbb R\}$. Let $P(r,s) = (r - s,0)$. Then $P$ is linear, $P^2 = P$, and $P(H) = V$, but if $x = (2,1)$, $$P(x) = (1,0)$$ while $$\operatorname{argmin}_{y\in V}\langle y - x, y - x \rangle^2 = (2,0)$$ and $$\operatorname{argmin}_{y\in ...


2

Your proposed domain for the adjoint $X^\star$ appears to me to be correct. As defined, $$ (Xf,g) = \sum_{j=0}^{\infty}(\sqrt{j+1}f_{j+1}+\sqrt{j}f_{j-1})\overline{g_j}. $$ By definition of adjoint, $g\in\mathcal{D}(X^{\star})$ iff there exists $h \in \ell^2$ such that the following holds for all $f \in \mathcal{D}(X)$: $$ ...


1

By definition, unitaries preserve the norm. So $$ \|Cy-CSy\|=\|C(y-Sy)\|=\|y-Sy\|. $$ Thus, $CV=V$.


1

Hint: We define $\varphi : V \to \mathbb R$ by $\varphi(v) := \langle f,v\rangle_{V^*,V}$. Now, we view $V$ as a (dense) subspace of $H$ and, thus, the functional $\varphi$ is defined on this dense subspace and continuous w.r.t. the norm in $H$.


1

You are correct. We can note that our space is isometric to the subspace of $l^2(\mathbb{N})$ comprised of sequences with finitely many non-zero entries, by the isometry $\sum_{n=0}^N a_n z^n\mapsto (a_0, a_1, \ldots, a_N, 0, 0, \ldots)$. (Note that, indeed, $\left\langle \sum_{n=0}^N a_n z^n, \sum_{n=0}^M b_n z^n\right\rangle = ...


2

$X=C[0,1]$ is not complete with respect to the norm $\|f\|=(f,f)^{1/2}=\int_0^1|f(t)|^2dt$. The completion of $C[0,1]$ with respect to this norm is $L^2[0,1]$. Every $f \in X$ can be written as $$ f = \left[f-\frac{(f,t^2)}{(t^2,t^2)}t^2\right]+\frac{(f,t^2)}{(t^2,t^2)}t^2 $$ The vector in square brackets is orthogonal to $t^2$. So ...


0

A metric space is complete iff every Cauchy sequence converges. If we want to prove that $X_0$ is incomplete, we have the advantage that all of our spaces are subspaces of the larger Hilbert space $L^2[0,1].$ Find a sequence in $X_0$ that converges (in $L^2$) to a discontinuous function. The sequence is Cauchy (that property only depends on the metric, not ...


1

a) The first task here is to interpret the question appropriately. I guess it should be read as, "Show that for every partial isometry $V\in B(H)$ on a finite-dimensional Hilbert space $H$, the restriction $V|_{(\ker V)^{\perp}}$ can be extended to a unitary on $H$." (The only extension of $V$ to $H$ is $V$ itself, and it is certainly not true that every ...


1

Assuming you see how to show it's a semigroup of bounded operators and you're just stuck on the "strongly continuous" part: We need to show that $||x-S(t)x||\to0$ as $t\to0$. Say $x=\sum a_ne_n$, where $\sum|a_n|^2<\infty$. Then $$||x-S(t)x||^2=\sum|1-e^{\lambda_nt}|^2|a_n|^2.$$The hypothesis on $\lambda_n$ shows that there exists $c$ with ...


1

The map $$T:L^2\to l^2:f\mapsto\left(a_n=\frac1{\sqrt{2\pi}}\int f(t)\exp(-int)dt\right)_{n\in\mathbb Z}$$ sends any square integrable function to the sequence of its Fourier coefficients, so we have $$\eqalignno{f(x)&=\sum_{n=-\infty}^{+\infty}a_n\exp(inx)&(1)}$$ for almost every $x$. The fact that $T$ is an isometry is known as Parseval's ...


2

Let $(z_n)$ be a convergent sequence in $L_1 + L_2$. Then we can write $z_n = \underbrace{x_n}_{\in L_1} +\underbrace{y_n}_{\in L_2}$. Since $L_1 \perp L_2 $ we have $\| z_n \|^2 = \|x_n \|^2 +\|y_n\|^2$ (Pythagoras). Moreover, $(z_n)$ is cauchy, and so: $$ \|x_n -x_m\|^2 +\|y_n -y_m\|^2 = \| z_n -z_m\|^2 \rightarrow 0$$ Implying $x_n\rightarrow x \space ; ...


2

The weak-* topology on $H^*$ is generated by the functionals $$\{h^* \mapsto h^*(x) \mid x \in H\}$$ If you pull-back this topology to $H$, you get the topology generated by $$\{y \mapsto \Phi(y)(x) \mid x \in H\} = \{y \mapsto (x,y) \mid x \in H\}$$ But this is the same as the weak topology which is generated by $$\{y \mapsto (y,x) \mid x \in H\}$$


1

The operator $A$ is closed. So $Y=\mathcal{D}(A)$ is a Hilbert space under the graph inner-product $$ (x,y)_A = (x,y)+(Ax,Ay). $$ This is the same as your form norm on $\mathcal{D}(A^{\star}A)$. Your question is equivalent to asking if $\mathcal{D}(A^\star A)$ is dense in $Y$. To prove that $\mathcal{D}(A^{\star}A)$ is dense in the form space, ...


1

There is a post here that can be useful.


1

Note that it is not a priori clear that a map satisfying $T(e^{inx}) = a_n$ exists. In fact, there exists infinitely many different linear maps $T \colon L^2([-\pi,\pi]) \rightarrow \ell_2(\mathbb{Z})$ but only one of those maps will be the map you want so it is not enough to say "we define $T$ by $T(e^{inx}) = a_n$". In linear algebra, given an (algebraic) ...


1

Fact 1. Every subnormal operator is Hyponormal. Fact 2. In finite dimension every Hyponormal operator is normal operator.(Since $tr(A^*A-AA^*)=0$ and only positive definite operator having trace zero is the zero operator.) So if you consider any non normal operator in finite dimensional Hilbert space, for example any nilpotent matrix, gives you a example ...


1

You are confused: Let $H$ be a Hilbert Space, let $B=\{u_j\}_{j=1}^\infty$ be a countable orthonormal basis. So we know that if a set is a complete orthonormal basis, the set of all finite linear combinations is dense in $H$. It is true. Now, since the set of all finite linear combinations is dense let $x\in H$, we have ...


0

If $A$ is positive, then the condition $\langle Ax,x\rangle=0$ implies $Ax=0$. Indeed, every positive operator $B$ on a Hilbert space admits a unique positive root $\sqrt B$ that satisfies $(\sqrt B)^2=B$. In your case this yields $$||\sqrt Ax\|^2=\langle \sqrt{A}x,\sqrt{A}x\rangle=\langle Ax,x\rangle=0.$$ Therefore we have $\sqrt Ax=0$ and finally ...


1

We may assume that $H_0=H$ since $\lVert \langle \cdot,h_n\rangle-\langle \cdot,h_0\rangle\rVert \to 0$ if $h_n\to h_0$, by dominated convergence. Take a function $f\in L^2(\mu)$ such that $\int fg \mathrm d\mu=0$ for each $g\in \mathcal C_0$. Decomposing $f$ into positive and negative parts ($f^+$ and $f^-$ respectively), we have for each $h_0\in H$, ...


1

$P_M$ must be the orthogonal projection. That means that $a-P_Ma$ is othogonal to $M$ and hence for any $x\in M$, we have $\|a-x\|^2=\|a-P_Ma\|^2+\|P_Ma-x\|^2\ge \|a-P_Ma\|^2$ (with equality iff $x=P_Ma$).


2

An aproach is using that every subnormal operator is hyponormal. Then, if we exhibit an non hyponormal, we finish. Take $H=\ell^2$ and $S$ the right shift. $T=(S^\ast+2S)^2$ is not hyponormal (the prove is straighforward). Thus $T$ is not subnormal.


2

Let $\rho = \sup_{\|x\|=1}\Re (Tx,x)$. Suppose $\rho <\Re\lambda$. Then $\Re(Tx,x)\le \rho(x,x)$ for all $x$, and \begin{align} (\Re\lambda-\rho)\|x\|^2 &\le \Re ((\lambda I-T)x,x) \\ (\Re\lambda-\rho)\|x\|^2 &\le \|(\lambda I-T)x\|\|x\| \\ (\Re\lambda-\rho)\|x\| &\le \|(\lambda I-T)x\|. \end{align} Therefore ...


0

The direct sum $H=E\oplus \mathbb C$ (or $\mathbb R$) define an hyperplan $E$. This $E$ is a maximal subespace of $H$ and is the kernel of a linear form $f:H\to \mathbb C$ (or $\mathbb R$) i. e. $E=f^{-1}(\{0\})$. A typical case of subspace dense is given when $f$ is not continuous: in fact $\bar E$ is still a subspace of $H$ (by continuity of the sum and ...


1

Not quite; it means that for any element $x$ in the Hilbert Space and for any $\epsilon > 0$, there exists an element $s$ in the span such that $$|x-s| < \epsilon$$


1

A subset $A$ of a topological (or in particular metric) space $X$ is dense if its closure is $X.$ In a metric space (and a Hilbert space is a metric space) this is equivalent to the condition that every element of $X$ is the limit of some convergent sequence with its elements in $A.$ The word "span" is already contained in "finite linear combination of ...


1

Yes. Simply note $$ \|AB \phi - BA \phi \| \leq \|AB \phi - A_n B \phi\| + \|A_n B\phi - BA\phi\| = \|AB \phi- A_n B \phi\| + \|B A_n \phi - BA\phi\|\leq \|B\phi\|\cdot \|A-A_n\| + \|B\| \|A_n \phi - A\phi\| $$ as $n\to \infty$. Here, I assumed $A_n \to A$ in operator norm. But actually, an easy modification of the argument shows that it suffices to have ...


1

What you do is to define $\phi $ by linearity on $$\mathcal M_\phi=\text {span}\,\{a\in A^+:\ \phi (a)<\infty\}. $$ Then the inequality from your comment shows that $b^*a\in \mathcal M_\phi $ whenever $a,b\in\mathcal N_\phi $.


2

In my experience, the topic of subalgebras of $M_n(\mathbb C)$ is not part of the usual linear algebra curriculum. What you need to understand first is the form that finite-dimensional C$^*$-algebras have. A finite-dimensional C$^*$-algebra $A$ is always a finite direct sum $$\bigoplus_{k=1}^m M_{n(k)}(\mathbb C).$$ The "blocks" can be identified via the ...


1

As you pointed out in your comment, $\hat{A} : \hat{H}\rightarrow\hat{H}$ is bounded. Therefore $\hat{A}-\lambda I$ is invertible for $|\lambda| > \|\hat{A}\|_{\hat{H}}$, which guarantees that $\mathcal{R}(\hat{A}-\lambda I)=\hat{H}$ for such $\lambda$, or $(A-\lambda I)\mathcal{D}(A)=\mathcal{D}(A)$ for such $\lambda$.


1

To show that $\{ \alpha_n(x)\beta_m(y) \}_{n,m}$ is an orthonormal basis, suppose that $f\in L^2(\mathbb{R}^{p}\times\mathbb{R}^{q})$ satisfies $\int_{\mathbb{R}^{p+q}}\alpha_n(x)\beta_m(y)f(x,y)dxdy = 0$ for all $n,m$. It is shown that $f=0$ a.e.. Because $\alpha_n(x)\beta_m(y)f(x,y) \in L^1(\mathbb{R}^{p+q})$, Fubini's Theorem for complete $\sigma$-finite ...


1

An answer that is equivalent to your friend's answer, but without technically invoking the concept of differentiation on normed linear spaces: For $x\in H$, consider the function $h_x:\mathbb{R}\rightarrow\mathbb{R}$ defined by $h_x(t) = g(u+tx)$. Then $h_x$ has a minimum at $t=0$, so if $h_x$ is differentiable there, then $h_x'(0)=0$, i.e. if the limit ...



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